JEE Main 202311 Apr, Evening shiftMathematicsMediumMatrices and Determinants245× since 2002∣x+1xxxx+λxxxx+λ2∣=98(103x+81)\left|\begin{array}{ccc}x+1 & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda^{2}\end{array}\right|=\frac{9}{8}(103 x+81)x+1xxxx+λxxxx+λ2=89(103x+81), then λ,λ3\lambda, \frac{\lambda}{3}λ,3λ are the roots of the equation :AOption A. 4x2+24x−27=04 x^{2}+24 x-27=04x2+24x−27=0BOption B. 4x2−24x+27=04 x^{2}-24 x+27=04x2−24x+27=0COption C. 4x2−24x−27=04 x^{2}-24 x-27=04x2−24x−27=0DOption D. 4x2+24x+27=04 x^{2}+24 x+27=04x2+24x+27=0CheckSkip · show solution
